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Sunday Times Teaser 2975 – Hollow Cube Land

by Bill Kinally

Published September 29 2019 (link)

I have a large box of toy building bricks. The bricks are all cubes (the same size), and can be pushed together then dismantled.

I decided to build the largest cube possible by leaving out all the interior bricks. When my hollow cube was finished I had two bricks left over. I put all the bricks back in the box and gave it to my two children. Each in turn was able to use every brick in the box to construct two hollow cubes, again with all interior bricks removed. Their cubes were all different sizes.

I told them this would not have been possible had the box contained any fewer bricks.

How many bricks were in the box?

2 Comments Leave one →
  1. Brian Gladman permalink

    If the large hollow cube has edges of \(r\) bricks, the number of bricks in its outer layer is given by \[r^3 – (r – 2)^3=6r^2+12r+8=6(r – 1)^2 + 2\] If the two smaller cubes are of side \(p\) and \(q\) we then have:\[\{6(r-1)^2 + 2\} + 2 = \{6(p-1)^2 + 2\} + \{6(q-1)^2 + 2\}\] which reduces to \[(p-1)^2+(q-1)^2=(r-1)^2\] which is a Pythagorean triple. Since the boys produce different pairs of hollow cubes from the same number of bricks, we need the smallest pair of Pythagorean triples that share a common hypotenuse. This can be easily obtained by inspection from this table, giving \(r-1 = 25\) and hence \(3754\) as the number of bricks.

    Here is a programmed solution:

  2. Erling Torkildsen permalink

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